Tighter constraints of multiqubit entanglement

TL;DR

This paper introduces tighter monogamy and polygamy inequalities for multiqubit entanglement using various entanglement measures, enhancing the understanding of entanglement distribution in quantum systems.

Contribution

It presents new classes of inequalities that are tighter than previous ones, applicable to different entanglement measures in multiqubit systems.

Findings

Inequalities are tighter for certain quantum states.

Applicable to multiple entanglement measures.

Improve understanding of entanglement distribution.

Abstract

Monogamy and polygamy relations characterize the distributions of entanglement in multipartite systems. We provide classes of monogamy and polygamy inequalities of multiqubit entanglement in terms of concurrence, entanglement of formation, negativity, Tsallis-$q$ entanglement and R\'{e}nyi-$\alpha$ entanglement, respectively. We show that these inequalities are tighter than the existing ones for some classes of quantum states.

Equations

$$\mathcal{E}({\rho }_{A|BC})\ge \mathcal{E}({\rho }_{AB})+\mathcal{E}({\rho }_{AC}),$$

$${\mathcal{E}}^a({\rho }_{A|BC})\le {\mathcal{E}}^a({\rho }_{AB})+{\mathcal{E}}^a({\rho }_{AC}),$$

$$\begin{array}[]{rl}&\mathcal{C}^{\beta}(\rho_{A|B_{1}\cdots B_{N-1}})\\ &\ \ \geq\mathcal{C}^{\beta}(\rho_{AB_{1}})+\mathcal{C}^{\beta}(\rho_{AB_{2}})+\cdots+\mathcal{C}^{\beta}(\rho_{AB_{N-1}}),\end{array}$$

$$\begin{array}[]{rl}&\mathcal{C}^{\beta}(\rho_{A|B_{1}\cdots B_{N-1}})\\ &\ \ \geq\mathcal{C}^{\beta}(\rho_{AB_{1}})+\frac{\beta}{2}\mathcal{C}^{\beta}(\rho_{AB_{2}})+\cdots\\ &\ \ \ \ +\big{(}\frac{\beta}{2}\big{)}^{m-1}\mathcal{C}^{\beta}(\rho_{AB_{m}})\\ &\ \ \ \ +\big{(}\frac{\beta}{2}\big{)}^{m+1}[\mathcal{C}^{\beta}(\rho_{AB_{m+1}})+\cdots+\mathcal{C}^{\beta}(\rho_{AB_{N-2}})]\\ &\ \ \ \ +\big{(}\frac{\beta}{2}\big{)}^{m}\mathcal{C}^{\beta}(\rho_{AB_{N-1}})\end{array}$$

$$\begin{array}[]{rl}&\mathcal{C}^{\beta}(\rho_{A|B_{1}\cdots B_{N-1}})\\ &\ \ \geq\mathcal{C}^{\beta}(\rho_{AB_{1}})+\big{(}2^{\frac{\beta}{2}}-1\big{)}\mathcal{C}^{\beta}(\rho_{AB_{2}})+\cdots\\ &\ \ \ \ +\big{(}2^{\frac{\beta}{2}}-1\big{)}^{m-1}\mathcal{C}^{\beta}(\rho_{AB_{m}})\\ &\ \ \ \ +\big{(}2^{\frac{\beta}{2}}-1\big{)}^{m+1}[\mathcal{C}^{\beta}(\rho_{AB_{m+1}})+\cdots+\mathcal{C}^{\beta}(\rho_{AB_{N-2}})]\\ &\ \ \ \ +\big{(}2^{\frac{\beta}{2}}-1\big{)}^{m}\mathcal{C}^{\beta}(\rho_{AB_{N-1}})\end{array}$$

$${\mathcal{C}}_a({\rho }_{AB})=\underset{\{p_i,|{\psi }_i⟩\}}{\max }\sum _i{p}_i\mathcal{C}(|{\psi }_i⟩),$$

$$\begin{array}[]{rl}&\mathcal{C}^{2}(|\psi\rangle_{A|B_{1}\cdots B_{N-1}})\\ &\ \ =\mathcal{C}_{a}^{2}(|\psi\rangle_{A|B_{1}\cdots B_{N-1}})\\ &\ \ \leq\mathcal{C}_{a}^{2}(\rho_{AB_{1}})+\mathcal{C}_{a}^{2}(\rho_{AB_{2}})+\cdots+\mathcal{C}_{a}^{2}(\rho_{AB_{N-1}}).\end{array}$$

$$(1+t{)}^m\ge 1+\frac{(1+k{)}^m-1}{k^m}{t}^m$$

$$(1+t{)}^n\le 1+\frac{(1+k{)}^n-1}{k^n}{t}^n$$

$$f(m,x)\geq f(m,\frac{1}{k})=\big{(}1+\frac{1}{k}\big{)}^{m}-\big{(}\frac{1}{k}\big{)}^{m}=\frac{(k+1)^{m}-1}{k^{m}}.$$

$${\mathcal{C}}_{A|BC}^{\beta }\ge {\mathcal{C}}_{AB}^{\beta }+\frac{(1+k{)}^{\frac{\beta }{2}}-1}{k^{\frac{\beta }{2}}}{\mathcal{C}}_{AC}^{\beta },$$

$$\begin{array}[]{rl}&\mathcal{C}_{A|BC}^{\beta}\geq(\mathcal{C}_{AB}^{2}+\mathcal{C}_{AC}^{2})^{\frac{\beta}{2}}\\[4.2679pt] &\ \ \ \ \ \ \ \ \ =\mathcal{C}_{AB}^{\beta}\Big{(}1+\frac{\mathcal{C}_{AC}^{2}}{\mathcal{C}_{AB}^{2}}\Big{)}^{\frac{\beta}{2}}\\[4.2679pt] &\ \ \ \ \ \ \ \ \ \geq\mathcal{C}_{AB}^{\beta}\Big{[}1+\frac{(1+k)^{\frac{\beta}{2}}-1}{k^{\frac{\beta}{2}}}\Big{(}\frac{\mathcal{C}_{AC}^{2}}{\mathcal{C}_{AB}^{2}}\Big{)}^{\frac{\beta}{2}}\Big{]}\\[8.53581pt] &\ \ \ \ \ \ \ \ \ =\mathcal{C}_{AB}^{\beta}+\frac{(1+k)^{\frac{\beta}{2}}-1}{k^{\frac{\beta}{2}}}\mathcal{C}_{AC}^{\beta},\end{array}$$

$$\begin{array}[]{rl}&\mathcal{C}^{\beta}_{A|B_{1}\cdots B_{N-1}}\\[5.69054pt] &\ \ \geq\mathcal{C}^{\beta}_{AB_{1}}+\frac{(1+k)^{\frac{\beta}{2}}-1}{k^{\frac{\beta}{2}}}\mathcal{C}^{\beta}_{AB_{2}}+\cdots\\[5.69054pt] &\ \ \ \ +\Big{(}\frac{(1+k)^{\frac{\beta}{2}}-1}{k^{\frac{\beta}{2}}}\Big{)}^{m-1}\mathcal{C}^{\beta}_{AB_{m}}\\[5.69054pt] &\ \ \ \ +\Big{(}\frac{(1+k)^{\frac{\beta}{2}}-1}{k^{\frac{\beta}{2}}}\Big{)}^{m+1}\Big{(}\mathcal{C}^{\beta}_{AB_{m+1}}+\cdots+\mathcal{C}^{\beta}_{AB_{N-2}}\Big{)}\\[5.69054pt] &\ \ \ \ +\Big{(}\frac{(1+k)^{\frac{\beta}{2}}-1}{k^{\frac{\beta}{2}}}\Big{)}^{m}\mathcal{C}^{\beta}_{AB_{N-1}}\end{array}$$

$$\begin{array}[]{rl}&\mathcal{C}_{A|B_{1}B_{2}\cdots B_{N-1}}\\[5.69054pt] &\ \ \geq\mathcal{C}_{AB_{1}}^{\beta}+\frac{(1+k)^{\frac{\beta}{2}}-1}{k^{\frac{\beta}{2}}}\mathcal{C}_{A|B_{2}\cdots B_{N-1}}^{\beta}\\[5.69054pt] &\ \ \geq\mathcal{C}_{AB_{1}}^{\beta}+\frac{(1+k)^{\frac{\beta}{2}}-1}{k^{\frac{\beta}{2}}}\mathcal{C}_{AB_{2}}^{\beta}\\[5.69054pt] &\ \ \ \ +\Big{(}\frac{(1+k)^{\frac{\beta}{2}}-1}{k^{\frac{\beta}{2}}}\Big{)}^{2}\mathcal{C}_{A|B_{3}\cdots B_{N-1}}^{\beta}\\[5.69054pt] &\ \ \geq\cdots\\[5.69054pt] &\ \ \geq\mathcal{C}_{AB_{1}}^{\beta}+\Big{(}\frac{(1+k)^{\frac{\beta}{2}}-1}{k^{\frac{\beta}{2}}}\Big{)}\mathcal{C}_{AB_{2}}^{\beta}\\[5.69054pt] &\ \ \ \ +\cdots+\Big{(}\frac{(1+k)^{\frac{\beta}{2}}-1}{k^{\frac{\beta}{2}}}\Big{)}^{m-1}\mathcal{C}_{AB_{m}}^{\beta}\\[5.69054pt] &\ \ \ \ +\Big{(}\frac{(1+k)^{\frac{\beta}{2}}-1}{k^{\frac{\beta}{2}}}\Big{)}^{m}\mathcal{C}_{A|B_{m+1}\cdots B_{N-1}}^{\beta}.\end{array}$$

$$\begin{array}[]{rl}&\mathcal{C}_{A|B_{m+1}\cdots B_{N-1}}^{\beta}\\[5.69054pt] &\ \ \geq\frac{(1+k)^{\frac{\beta}{2}}-1}{k^{\frac{\beta}{2}}}\mathcal{C}_{AB_{m+1}}^{\beta}+\mathcal{C}_{A|B_{m+2\cdots B_{N-1}}}^{\beta}\\[5.69054pt] &\ \ \geq\frac{(1+k)^{\frac{\beta}{2}}-1}{k^{\frac{\beta}{2}}}\Big{(}\mathcal{C}_{AB_{m+1}}^{\beta}+\cdots+\mathcal{C}_{AB_{N-2}}^{\beta}\Big{)}+\mathcal{C}_{AB_{N-1}}^{\beta}.\end{array}$$

$$\begin{array}[]{rcl}\mathcal{C}^{\beta}_{A|B_{1}\cdots B_{N-1}}&\geq&\mathcal{C}^{\beta}_{AB_{1}}+\frac{(1+k)^{\frac{\beta}{2}}-1}{k^{\frac{\beta}{2}}}\mathcal{C}^{\beta}_{AB_{2}}+\cdots\\[8.53581pt] &&+\Big{(}\frac{(1+k)^{\frac{\beta}{2}}-1}{k^{\frac{\beta}{2}}}\Big{)}^{N-2}\mathcal{C}^{\beta}_{AB_{N-1}}\end{array}$$

$$\frac{(1+k{)}^{\frac{\beta }{2}}-1}{k^{\frac{\beta }{2}}}\ge 2^{\frac{\beta }{2}}-1$$

$$|\psi {⟩}_{ABC}={\lambda }_0|000⟩+{\lambda }_1{e}^{i\phi }|100⟩+{\lambda }_2|101⟩+{\lambda }_3|110⟩+{\lambda }_4|111⟩,$$

$$\begin{array}[]{rl}&\mathcal{C}_{a}^{\beta}(|\psi\rangle_{A|B_{1}\cdots B_{N-1}})\\[5.69054pt] &\ \ \leq\mathcal{C}^{\beta}_{aAB_{1}}+\frac{(1+k)^{\frac{\beta}{2}}-1}{k^{\frac{\beta}{2}}}\mathcal{C}^{\beta}_{aAB_{2}}+\cdots\\[5.69054pt] &\ \ \ \ +\Big{(}\frac{(1+k)^{\frac{\beta}{2}}-1}{k^{\frac{\beta}{2}}}\Big{)}^{m-1}\mathcal{C}^{\beta}_{aAB_{m}}\\[5.69054pt] &\ \ \ \ +\Big{(}\frac{(1+k)^{\frac{\beta}{2}}-1}{k^{\frac{\beta}{2}}}\Big{)}^{m+1}\Big{(}\mathcal{C}^{\beta}_{aAB_{m+1}}+\cdots+\mathcal{C}^{\beta}_{aAB_{N-2}}\Big{)}\\[5.69054pt] &\ \ \ \ +\Big{(}\frac{(1+k)^{\frac{\beta}{2}}-1}{k^{\frac{\beta}{2}}}\Big{)}^{m}\mathcal{C}^{\beta}_{aAB_{N-1}}\end{array}$$

$$\begin{array}[]{rcl}\mathcal{C}^{\beta}_{a}{|\psi\rangle_{A|B_{1}\cdots B_{N-1}}}&\leq&\mathcal{C}^{\beta}_{aAB_{1}}+\frac{(1+k)^{\frac{\beta}{2}}-1}{k^{\frac{\beta}{2}}}\mathcal{C}^{\beta}_{aAB_{2}}+\cdots\\[8.53581pt] &&+\Big{(}\frac{(1+k)^{\frac{\beta}{2}}-1}{k^{\frac{\beta}{2}}}\Big{)}^{N-2}\mathcal{C}^{\beta}_{aAB_{N-1}}\end{array}$$

$$E(|\psi {⟩}_{AB})=\mathcal{S}({\rho }_A),$$

$$E({\rho }_{AB})=\underset{\{p_i,|{\psi }_i⟩\}}{\min }\sum _i{p}_{i}E(|{\psi }_i⟩),$$

$$f^{\sqrt{2}}(x^2+y^2)\ge f^{\sqrt{2}}(x^2)+f^{\sqrt{2}}(y^2),$$

$$E_{A|B_1{B}_2\cdots B_{N-1}}^{\beta }\ge E_{A{B}_1}^{\beta }+E_{A{B}_2}^{\beta }+\cdots +E_{A{B}_{N-1}}^{\beta },$$

$$\begin{array}[]{rl}&E^{\beta}_{A|B_{1}B_{2}\cdots B_{N-1}}\\[4.2679pt] &\ \ \geq E^{\beta}_{AB_{1}}+\frac{\beta}{\sqrt{2}}E^{\beta}_{AB_{2}}+\cdots+\big{(}\frac{\beta}{\sqrt{2}}\big{)}^{m-1}E^{\beta}_{AB_{m}}\\[4.2679pt] &\ \ \ \ +\big{(}\frac{\beta}{\sqrt{2}}\big{)}^{m+1}\big{(}E^{\beta}_{AB_{m+1}}+\cdots+E^{\beta}_{AB_{N-2}}\big{)}\\[4.2679pt] &\ \ \ \ +\big{(}\frac{\beta}{\sqrt{2}}\big{)}^{m}E^{\beta}_{AB_{N-1}},\end{array}$$

$$\begin{array}[]{rl}&E^{\beta}_{A|B_{1}B_{2}\cdots B_{N-1}}\\[4.2679pt] &\ \ \geq E^{\beta}_{AB_{1}}+\Big{(}2^{\frac{\beta}{\sqrt{2}}}-1\Big{)}E^{\beta}_{AB_{2}}+\cdots+\Big{(}2^{\frac{\beta}{\sqrt{2}}}-1\Big{)}^{m-1}\\[4.2679pt] &\ \ \ \ \times E^{\beta}_{AB_{m}}+\Big{(}2^{\frac{\beta}{\sqrt{2}}}-1\Big{)}^{m+1}\big{(}E^{\beta}_{AB_{m+1}}+\cdots+E^{\beta}_{AB_{N-2}}\big{)}\\[4.2679pt] &\ \ \ \ +\Big{(}2^{\frac{\beta}{\sqrt{2}}}-1\Big{)}^{m}E^{\beta}_{AB_{N-1}},\end{array}$$

$$\begin{array}[]{rl}&E^{\beta}_{A|B_{1}B_{2}\cdots B_{N-1}}\\[5.69054pt] &\ \ \geq E^{\beta}_{AB_{1}}+\frac{(1+k)^{t}-1}{k^{t}}E^{\beta}_{AB_{2}}+\cdots+\Big{(}\frac{(1+k)^{t}-1}{k^{t}}\Big{)}^{m-1}E^{\beta}_{AB_{m}}\\[5.69054pt] &\ \ \ \ +\Big{(}\frac{(1+k)^{t}-1}{k^{t}}\Big{)}^{m+1}(E^{\beta}_{AB_{m+1}}+\cdots+E^{\beta}_{AB_{N-2}})\\[5.69054pt] &\ \ \ \ +\Big{(}\frac{(1+k)^{t}-1}{k^{t}}\Big{)}^{m}E^{\beta}_{AB_{N-1}},\end{array}$$

$$\begin{array}[]{rl}&f^{\beta}(x^{2}+y^{2})=[f^{\sqrt{2}}(x^{2}+y^{2})]^{t}\\[4.2679pt] &\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \geq[f^{\sqrt{2}}(x^{2})+f^{\sqrt{2}}(y^{2})]^{t}\\[4.2679pt] &\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \geq[f^{\sqrt{2}}(x^{2})]^{t}+\frac{(1+k)^{t}-1}{k^{t}}[f^{\sqrt{2}}(y^{2})]^{t}\\[4.2679pt] &\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =f^{\beta}(x^{2})+\frac{(1+k)^{t}-1}{k^{t}}f^{\beta}(y^{2}),\end{array}$$

$$E_{A|B_1{B}_2\cdots B_{N-1}}\ge f({\mathcal{C}}_{A|B_1{B}_2\cdots B_{N-1}}^2).$$

$$\begin{array}[]{rl}&E^{\beta}_{A|B_{1}B_{2}\cdots B_{N-1}}\\[5.69054pt] &\ \ \geq f^{\beta}(\mathcal{C}^{2}_{A|B_{1}B_{2}\cdots B_{N-1}})\\[5.69054pt] &\ \ \geq f^{\beta}(\mathcal{C}^{2}_{A|B_{1}})+\frac{(1+k)^{t}-1}{k^{t}}f^{\beta}(\mathcal{C}^{2}_{A|B_{2}})+\cdots\\[5.69054pt] &\ \ \ \ +\Big{(}\frac{(1+k)^{t}-1}{k^{t}}\Big{)}^{m-1}f^{\beta}(\mathcal{C}^{2}_{A|B_{m}})+\Big{(}\frac{(1+k)^{t}-1}{k^{t}}\Big{)}^{m+1}\\[5.69054pt] &\ \ \ \ \ [f^{\beta}(\mathcal{C}^{2}_{AB_{m+1}})+\cdots+f^{\beta}(\mathcal{C}^{2}_{AB_{N-2}})]\\[5.69054pt] &\ \ \ \ +\Big{(}\frac{(1+k)^{t}-1}{k^{t}}\Big{)}^{m}f^{\beta}(\mathcal{C}^{2}_{AB_{N-1}})\\[5.69054pt] &\ \ =E^{\beta}_{AB_{1}}+\Big{(}\frac{(1+k)^{t}-1}{k^{t}}\Big{)}E^{\beta}_{AB_{2}}+\cdots+\Big{(}\frac{(1+k)^{t}-1}{k^{t}}\Big{)}^{m-1}\\[5.69054pt] &\ \ \ \ \times E^{m-1}_{AB_{m}}+\Big{(}\frac{(1+k)^{t}-1}{k^{t}}\Big{)}^{m+1}(E^{\beta}_{AB_{m+1}}+\cdots+E^{\beta}_{AB_{N-2}})\\[5.69054pt] &\ \ \ \ +\Big{(}\frac{(1+k)^{t}-1}{k^{t}}\Big{)}^{m}E^{\beta}_{AB_{N-1}},\end{array}$$

Full text

I Introduction

Quantum entanglement is an essential feature of quantum mechanics which distinguishes the quantum from the classical world and plays a very important role in quantum information processing dynamics ; GaoYang ; demonstration ; QIp . One singular property of quantum entanglement is that a quantum system entangled with one of the other subsystems limits its entanglement with the remaining ones, known as the monogamy of entanglement (MoE) BMT ; JSK . MoE plays a key role in many quantum information and communication processing tasks such as the security proof in quantum cryptographic scheme CHB and the security analysis of quantum key distribution Pawl .

For a tripartite quantum state $\rho_{ABC}$, MoE can be described as the following inequality

[TABLE]

where $\rho_{AB}={\rm tr}_{C}(\rho_{ABC})$ and $\rho_{AC}={\rm tr}_{B}(\rho_{ABC})$ are reduced density matrices, and $\mathcal{E}$ is an entanglement measure. However, it has been shown that not all entanglement measures satisfy such monogamy relations. It has been shown that the squared concurrence $\mathcal{C}^{2}$ TJO ; Bai , the squared entanglement of formation (EoF) $E^{2}$ Oliveira and the squared convex-roof extended negativity (CREN) $\mathcal{N}_{c}^{2}$ JSK8 ; Feng1 satisfy the monogamy relations for multiqubit states.

Another important concept is the assisted entanglement, which is a dual amount to bipartite entanglement measure. It has a dually monogamous property in multipartite quantum systems and gives rise to polygamy relations. For a tripartite state $\rho_{ABC}$, the usual polygamy relation is of the form,

[TABLE]

where $\mathcal{E}^{a}$ is the corresponding entanglement measure of assistance associated to $\mathcal{E}$. Such polygamy inequality has been deeply investigated in recent years, and was generalized to multiqubit systems and classes of higher-dimensional quantum systems JSK3 ; F.B ; G.G1 ; G.G2 ; JSK5 ; JSK6 ; JSK8 ; Song .

Recently, generalized classes of monogamy inequalities related to the $\beta$th power of entanglement measures were proposed. In Ref. SM.Fei1 ; SM.Fei3 , the authors proved that the squared concurrence and CREN satisfy the monogamy inequalities in multiqubit systems for $\beta\geq 2$. It has also been shown that the EoF satisfies monogamy relations when $\beta\geq\sqrt{2}$ SM.Fei1 ; SM.Fei2 ; SM.Fei3 . Besides, the Tsallis-$q$ entanglement and Rényi-$\alpha$ entanglement satisfy monogamy relations when $\beta\geq 1$ JSK2 ; JSK3 ; SM.Fei2 ; SM.Fei3 for some cases. Moreover, the corresponding polygamy relations have also been established G.G1 ; G.G2 ; JSK5 ; JSK9 ; Song ; Yongming .

In this paper, we investigate monogamy relations and polygamy relations in multiqubit systems. We provide tighter constraints of multiqubit entanglement than all the existing ones, thus give rise to finer characterizations of the entanglement distributions among the multiqubit systems.

II Tighter constraints related to concurrence

We first consider the monogamy inequalities and polygamy inequalities for concurrence. For a bipartite pure state $|\psi\rangle_{AB}$ in Hilbert space $H_{A}\otimes H_{B}$, the concurrence is defined as fide ; SM.Fei4 $\mathcal{C}(|\psi\rangle_{AB})=\sqrt{2(1-{\rm tr}\rho_{A}^{2})}$ with $\rho_{A}={\rm tr}_{B}|\psi\rangle_{AB}\langle\psi|$. The concurrence for a bipartite mixed state $\rho_{AB}$ is defined by the convex roof extension, $\mathcal{C}(\rho_{AB})=\min\limits_{\{p_{i},|\psi_{i}\rangle\}}\sum\limits_{i}p_{i}\mathcal{C}(|\psi_{i}\rangle)$, where the minimum is taken over all possible decompositions of $\rho_{AB}=\sum\limits_{i}p_{i}|\psi_{i}\rangle\langle\psi_{i}|$ with $\sum p_{i}=1$ and $p_{i}\geq 0$. For an $N$-qubit state $\rho_{AB_{1}\cdots B_{N-1}}\in H_{A}\otimes H_{B_{1}}\otimes\cdots\otimes H_{B_{N-1}}$, the concurrence $\mathcal{C}(\rho_{A|B_{1}\cdots B_{N-1}})$ of the state $\rho_{AB_{1}\cdots B_{N-1}}$ under bipartite partition $A$ and $B_{1}\cdots B_{N-1}$ satisfies SM.Fei1

[TABLE]

for $\beta\geq 2$, where $\rho_{AB_{j}}$ denote two-qubit reduced density matrices of subsystems $AB_{j}$ for $j=1,2,\ldots,N-1$. Later, the relation (3) is improved for the case $\beta\geq 2$ SM.Fei2 as

[TABLE]

conditioned that $\mathcal{C}(\rho_{AB_{i}})\geq\mathcal{C}(\rho_{A|B_{i+1}\cdots B_{N-1}})$ for $i=1,2,\ldots,m$, and $\mathcal{C}(\rho_{AB_{j}})\leq\mathcal{C}(\rho_{A|B_{j+1}\cdots B_{N-1}})$ for $j=m+1,\ldots,N-2$. The relation (4) is further improved for $\beta\geq 2$ as SM.Fei3

[TABLE]

with the same conditions as in (4).

For a tripartite state $|\psi\rangle_{ABC}$, the concurrence of assistance (CoA) is defined by EOS ; monogamy

[TABLE]

where the maximun is taken over all possible pure state decompositions of $\rho_{AB}$, and $\mathcal{C}(|\psi\rangle_{AB})=\mathcal{C}_{a}(|\psi\rangle_{AB})$. The generalized polygamy relation based on the concurrence of assistance was established in G.G1 ; G.G2

[TABLE]

These monogamy and polygamy relations for concurrence can be further tightened under some conditions. To this end, we first introduce the following lemma.

Lemma 1**.**

Suppose that $k$ is a real number satisfying $0<k\leq 1$, then for any $0\leq t\leq k$ and non-negative real numbers $m,n$, we have

[TABLE]

for $m\geq 1$, and

[TABLE]

for $0\leq n\leq 1$.

Proof: We first consider the function $f(m,x)=(1+x)^{m}-x^{m}$ with $x\geq\frac{1}{k}$ and $m\geq 1$. Then $f(m,x)$ is an increasing function of $x$, since $\frac{\partial f(m,x)}{\partial x}=m[(1+x)^{m-1}-x^{m-1}]\geq 0$. Thus,

[TABLE]

Set $x=\frac{1}{t}$ in (10), we get the inequality (8).

Similar to the proof of inequality (8), we can obtain the inequality (9), since in this case $f(n,x)$ is a decreasing function of $x$ for $x\geq\frac{1}{k}$ and $0\leq n\leq 1$.

In the next, we denote $\mathcal{C}_{AB_{i}}=\mathcal{C}(\rho_{AB_{i}})$ the concurrence of $\rho_{AB_{i}}$ and $\mathcal{C}_{A|B_{1}\cdots B_{N-1}}=\mathcal{C}(\rho_{A|B_{1}\cdots B_{N-1}})$ for convenience.

Lemma 2**.**

Suppose that $k$ is a real number satisfying $0<k\leq 1$. Then for any $2\otimes 2\otimes 2^{n-2}$ mixed state $\rho\in H_{A}\otimes H_{B}\otimes H_{C}$, if $\mathcal{C}_{AC}^{2}\leq k\mathcal{C}_{AB}^{2}$, we have

[TABLE]

for all $\beta\geq 2$.

Proof: Since $\mathcal{C}_{AC}^{2}\leq k\mathcal{C}_{AB}^{2}$ and $\mathcal{C}_{AB}>0$, we obtain

[TABLE]

where the first inequality is due to the fact, $\mathcal{C}_{A|BC}^{2}\geq\mathcal{C}_{AB}^{2}+\mathcal{C}_{AC}^{2}$ for arbitrary $2\otimes 2\otimes 2^{n-2}$ tripartite state $\rho_{ABC}$ TJO ; XJR and the second is due to Lemma 1. We can also see that if $\mathcal{C}_{AB}=0$, then $\mathcal{C}_{AC}=0$, and the lower bound becomes trivially zero.

For multiqubit systems, we have the following Theorems.

Theorem 1**.**

Suppose $k$ is a real number satisfying $0<k\leq 1$. For an $N$-qubit mixed state $\rho_{AB_{1}\cdots B_{N-1}}$, if $k\mathcal{C}_{AB_{i}}^{2}\geq\mathcal{C}_{A|B_{i+1}\cdots B_{N-1}}^{2}$ for $i=1,2,\ldots,m$, and $\mathcal{C}_{AB_{j}}^{2}\leq k\mathcal{C}_{A|B_{j+1}\cdots B_{N-1}}^{2}$ for $j=m+1,\ldots,N-2$, $\forall 1\leq m\leq N-3$, $N\geq 4$, then we have

[TABLE]

for all $\beta\geq 2$.

Proof: From the inequality (11), we have

[TABLE]

Since $\mathcal{C}_{AB_{j}}^{2}\leq k\mathcal{C}_{A|B_{j+1}\cdots B_{N-1}}^{2}$, for $j=m+1,\ldots,N-2$, we get

[TABLE]

Combining (14) and (15), we get the inequality (13).

If we replace the conditions $k\mathcal{C}_{AB_{i}}\geq\mathcal{C}_{A|B_{i+1}\cdots B_{N-1}}$ for $i=1,2,\ldots,m$, and $\mathcal{C}_{AB_{j}}^{2}\leq k\mathcal{C}_{A|B_{j+1}\cdots B_{N-1}}^{2}$ for $j=m+1,\ldots,N-2$, $\forall 1\leq m\leq N-3$, $N\geq 4$, in Theorem 1 by $k\mathcal{C}_{AB_{i}}^{2}\geq\mathcal{C}_{A|B_{i+1}\cdots B_{N-1}}^{2}$ for $i=1,2,\ldots,N-2$, then we have the following theorem.

Theorem 2**.**

Suppose $k$ is a real number satisfying $0<k\leq 1$. For an $N$-qubit mixed state $\rho_{AB_{1}\cdots B_{N-1}}$, if $k\mathcal{C}_{AB_{i}}^{2}\geq\mathcal{C}_{A|B_{i+1}\cdots B_{N-1}}^{2}$ for all $i=1,2,\ldots,N-2$, then we have

[TABLE]

for $\beta\geq 2$.

It can be seen that the inequalities (13) and (16) are tighter than the ones given in Ref. SM.Fei3 , since

[TABLE]

for $\beta\geq 2$ and $0<k\leq 1$. The equality holds when $k=1$. Namely, the result (5) given in SM.Fei3 are just special cases of ours for $k=1$. As $\frac{(1+k)^{\frac{\beta}{2}}-1}{k^{\frac{\beta}{2}}}$ is a decreasing function with respect to $k$ for $0<k\leq 1$ and $\beta\geq 2$, we find that the smaller $k$ is, the tighter the inequalities (11), (13) and (16) are.

Example 1    Consider the three-qubit state $|\psi\rangle_{ABC}$ in generalized Schmidt decomposition form Schmidt ; SM.Fei5 ,

[TABLE]

where $\lambda_{i}\geq 0$, $i=,1,2...,4$, and $\sum\limits_{i=0}^{4}\lambda_{i}^{2}=1$. Then we get $\mathcal{C}_{A|BC}=2\lambda_{0}\sqrt{\lambda_{2}^{2}+\lambda_{3}^{2}+\lambda_{4}^{2}}$, $\mathcal{C}_{AB}=2\lambda_{0}\lambda_{2}$ and $\mathcal{C}_{AC}=2\lambda_{0}\lambda_{3}$. Set $\lambda_{0}=\lambda_{3}=\frac{1}{2},\ \lambda_{2}=\frac{\sqrt{2}}{2}$ and $\lambda_{1}=\lambda_{4}=0$. We have $\mathcal{C}_{A|BC}=\frac{\sqrt{3}}{2}$, $\mathcal{C}_{AB}=\frac{\sqrt{2}}{2}$ and $\mathcal{C}_{AC}=\frac{1}{2}$. Then \mathcal{C}_{AB}^{\beta}+\big{(}2^{\frac{\beta}{2}}-1\big{)}\mathcal{C}_{AC}^{\beta}=\big{(}\frac{\sqrt{2}}{2}\big{)}^{\beta}+\big{(}2^{\frac{\beta}{2}}-1\big{)}\big{(}\frac{1}{2}\big{)}^{\beta} and \mathcal{C}_{AB}^{\beta}+\frac{(1+k)^{\frac{\beta}{2}}-1}{k^{\frac{\beta}{2}}}\mathcal{C}_{AC}^{\beta}=\big{(}\frac{\sqrt{2}}{2}\big{)}^{\beta}+\frac{(1+k)^{\frac{\beta}{2}}-1}{k^{\frac{\beta}{2}}}\big{(}\frac{1}{2}\big{)}^{\beta}. One can see that our result is better than the result (5) in SM.Fei3 for $\beta\geq 2$, hence better than (3) and (4) given in SM.Fei1 ; SM.Fei2 , see Fig. 1.

We now discuss the polygamy relations for the CoA of $\mathcal{C}_{a}(|\psi\rangle_{A|B_{1}\cdots B_{N-1}})$ for $0\leq\beta\leq 2$. We have the following Theorem.

Theorem 3**.**

Suppose $k$ is a real number satisfying $0<k\leq 1$. For an $N$-qubit pure state $|\psi\rangle_{AB_{1}\cdots B_{N-1}}$, if $k\mathcal{C}^{2}_{aA|B_{i}}\geq\mathcal{C}^{2}_{aA|B_{i+1}\cdots B_{N-1}}$ for $i=1,2,\ldots,m$, and $\mathcal{C}^{2}_{aA|B_{j}}\leq k\mathcal{C}^{2}_{aA|B_{j+1}\cdots B_{N-1}}$ for $j=m+1,\ldots,N-2$, $\forall 1\leq m\leq N-3$, $N\geq 4$, then we have

[TABLE]

for all $0\leq\beta\leq 2$.

Proof: The proof is similar to the proof of Theorem 1 by using inequality (9).

Theorem 4**.**

Suppose $k$ is a real number satisfying $0<k\leq 1$. For an $N$-qubit pure state $|\psi\rangle_{AB_{1}\cdots B_{N-1}}$, if $k\mathcal{C}^{2}_{aAB_{i}}\geq\mathcal{C}^{2}_{aA|B_{i+1}\cdots B_{N-1}}$ for all $i=1,2,\ldots,N-2$, then we have

[TABLE]

for $0\leq\beta\leq 2$.

The inequalities (18) and (19) are also upper bounds of $\mathcal{C}(|\psi\rangle_{A|B_{1}\cdots B_{N-1}})$ for pure state $|\psi\rangle_{AB_{1}\cdots B_{N-1}}$ since $\mathcal{C}(|\psi\rangle_{A|B_{1}\cdots B_{N-1}})=\mathcal{C}_{a}(|\psi\rangle_{A|B_{1}\cdots B_{N-1}})$.

III Tighter constraints relate to EoF

Let $H_{A}$ and $H_{B}$ be two Hilbert spaces with dimension $m$ and $n$ $(m\leq n)$, respectively. Then the entanglement of formation (EoF) CHB1 ; CHB2 is defined as follows: for a pure state $|\psi\rangle_{AB}\in H_{A}\otimes H_{B}$, the EoF is given by

[TABLE]

where $\rho_{A}={\rm Tr}_{B}(|\psi\rangle_{AB}\langle\psi|)$ and $\mathcal{S}(\rho)=-{\rm Tr}(\rho\log_{2}\rho)$. For a bipartite mixed state $\rho_{AB}\in H_{A}\otimes H_{B}$, the EoF is given by

[TABLE]

with the minimum taking over all possible pure state decomposition of $\rho_{AB}$.

In Ref. EoF , Wootters showed that $E(|\psi\rangle)=f(\mathcal{C}^{2}(|\psi\rangle))$ for $2\otimes m\ (m\geq 2)$ pure state $|\psi\rangle$, and $E(\rho)=f(\mathcal{C}^{2}(\rho))$ for two-qubit mixed state $\rho$, where f(x)=H\big{(}\frac{1+\sqrt{1-x}}{2}\big{)} and $H(x)=-x\log_{2}x-(1-x)\log_{2}(1-x)$. $f(x)$ is a monotonically increasing function for $0\leq x\leq 1$, and satisfies the following relations:

[TABLE]

where $f^{\sqrt{2}}(x^{2}+y^{2})=[f(x^{2}+y^{2})]^{\sqrt{2}}$.

Although EoF does not satisfy the inequality $E_{AB}+E_{AC}\leq E_{A|BC}$ CKW , the authors in BZYW showed that EoF is a monotonic function satisfying $E^{2}(\rho_{A|B_{1}B_{2}\cdots B_{N-1}})\geq\sum_{i=1}^{N-1}E^{2}(\rho_{AB_{i}})$. For $N$-qubit systems, one has SM.Fei1

[TABLE]

for $\beta\geq\sqrt{2}$, where $E_{A|B_{1}B_{2}\cdots B_{N-1}}$ is the EoF of $\rho$ under bipartite partition $A|B_{1}B_{2}\cdots B_{N-1}$, and $E_{AB_{i}}$ is the EoF of the mixed state $\rho_{AB_{i}}={\rm Tr}_{B_{1}\cdots B_{i-1},B_{i+1}\cdots B_{N-1}}(\rho)$ for $i=1,2,\ldots,N-1$. Recently, the authors in Ref. SM.Fei2 proposed a monogamy relation that is tighter than the inequality (23),

[TABLE]

if $\mathcal{C}_{AB_{i}}\geq\mathcal{C}_{A|B_{j+1}\cdots B_{N-1}}$ for $i=1,2,\ldots,m$, and $\mathcal{C}_{AB_{j}}\leq\mathcal{C}_{A|B_{j+1}\cdots B_{N-1}}$ for $j=m+1,\ldots,N-2$, $\forall 1\leq m\leq N-3$, $N\geq 4$ for $\beta\geq\sqrt{2}$. The inequality (24) is also improved to

[TABLE]

under the same conditions as that of inequality (24).

In fact, these inequalities can be further improved to even tighter monogamy relations.

Theorem 5**.**

Suppose $k$ is a real number satisfying $0<k\leq 1$. For any $N$-qubit mixed state $\rho_{AB_{1}\cdots B_{N-1}}$, if $kE^{\sqrt{2}}_{AB_{i}}\geq E^{\sqrt{2}}_{A|B_{i+1}\cdots B_{N-1}}$ for $i=1,2,\ldots,m$, and $E^{\sqrt{2}}_{AB_{j}}\leq kE^{\sqrt{2}}_{A|B_{j+1}\cdots B_{N-1}}$ for $j=m+1,\ldots,N-2$, $\forall 1\leq m\leq N-3$, $N\geq 4$, the entanglement of formation $E(\rho)$ satisfies

[TABLE]

for $\beta\geq\sqrt{2}$, where $t=\frac{\beta}{\sqrt{2}}$.

Proof: For $\beta\geq\sqrt{2}$ and $kf^{\sqrt{2}}(x^{2})\geq f^{\sqrt{2}}(y^{2})$, we find

[TABLE]

where the first inequality is due to the inequality (22), and the second inequality can be obtained from inequality (8).

Let $\rho=\sum_{i}p_{i}|\psi_{i}\rangle\langle\psi_{i}|\in H_{A}\otimes H_{B_{1}}\otimes\cdots H_{B_{N-1}}$ be the optimal decomposition of $E_{A|B_{1}B_{2}\cdots B_{N-1}}(\rho)$ for the $N$-qubit mixed state $\rho$. Then SM.Fei3

[TABLE]

Thus,

[TABLE]

where the first inequality holds due to (28), the second inequality is similar to the proof of Theorem 1 by using inequality (27), and the last equality holds since for any $2\otimes 2$ quantum state $\rho_{AB_{i}}$, $E(\rho_{AB_{i}})=f[\mathcal{C}^{2}(\rho_{AB_{i}})]$.

Similar to the case of concurrence, we have also the following tighter monogamy relation for EoF.

Theorem 6**.**

Suppose $k$ is a real number satisfying $0<k\leq 1$. For an $N$-qubit mixed state $\rho_{AB_{1}\cdots B_{N-1}}$, if $kE^{\sqrt{2}}_{AB_{i}}\geq E^{\sqrt{2}}_{A|B_{i+1}\cdots B_{N-1}}$ for all $i=1,2,\ldots,N-2$, we have

[TABLE]

for $\beta\geq\sqrt{2}$ and $t=\frac{\beta}{\sqrt{2}}$.

As $\frac{(1+k)^{t}-1}{k^{t}}\geq 2^{t}-1$ for $t\geq 1$ and $0<k\leq 1$, our new monogamy relations (26) and (30) are tighter than the ones given in SM.Fei1 ; SM.Fei2 ; SM.Fei3 . Also, for $0<k\leq 1$ and $\beta\geq 2$, the smaller $k$ is, the tighter inequalities (26) and (30) are.

Example 2    Let us again consider the three-qubit state $|\psi\rangle_{ABC}$ defined in (17) with $\lambda_{0}=\lambda_{3}=\frac{1}{2},\ \lambda_{2}=\frac{\sqrt{2}}{2}$ and $\lambda_{1}=\lambda_{4}=0$. Then $E_{A|BC}=2-\log_{2}3\approx 0.811278$, $E_{AB}=-\frac{2+\sqrt{2}}{4}\log_{2}\frac{2+\sqrt{2}}{4}-\frac{2-\sqrt{2}}{4}\log_{2}\frac{2-\sqrt{2}}{4}\approx 0.600876$ and $E_{AB}=-\frac{2+\sqrt{3}}{4}\log_{2}\frac{2+\sqrt{3}}{4}-\frac{2-\sqrt{3}}{4}\log_{2}\frac{2-\sqrt{3}}{4}\approx 0.354579$. Thus, E^{\beta}_{AB}+\big{(}2^{\frac{\beta}{2}}-1\big{)}E^{\beta}_{AC}=(0.600876)^{\beta}+\big{(}2^{\frac{\beta}{2}}-1\big{)}0.354579^{\beta}, $E^{\beta}_{AB}+\frac{1.5^{\frac{\beta}{2}}-1}{0.5^{\frac{\beta}{2}}}E^{\beta}_{AC}=(0.600876)^{\beta}+\frac{1.5^{\frac{\beta}{2}}-1}{0.5^{\frac{\beta}{2}}}0.354579^{\beta}$ , $E^{\beta}_{AB}+\frac{1.7^{\frac{\beta}{2}}-1}{0.7^{\frac{\beta}{2}}}E^{\beta}_{AC}=(0.600876)^{\beta}+\frac{1.7^{\frac{\beta}{2}}-1}{0.7^{\frac{\beta}{2}}}0.354579^{\beta}$ and $E^{\beta}_{AB}+\frac{1.9^{\frac{\beta}{2}}-1}{0.9^{\frac{\beta}{2}}}E^{\beta}_{AC}=(0.600876)^{\beta}+\frac{1.9^{\frac{\beta}{2}}-1}{0.9^{\frac{\beta}{2}}}0.354579^{\beta}$. One can see that our result is better than the one in SM.Fei3 for $\beta\geq\sqrt{2}$, hencee better than the ones in SM.Fei1 ; SM.Fei2 , see Fig. 2.

We can also provide tighter polygamy relations for the entanglement of assistance. The entanglement of assistance (EoA) of $\rho_{AB}$ is defined as cohen ,

[TABLE]

with the maximization taking over all possible pure decompositions of $\rho_{AB}$. For any dimensional multipartite quantum state $\rho_{AB_{1}B_{2}\cdots B_{N-1}}$, a general polygamy inequality of multipartite quantum entanglement was established as JSK5 ,

[TABLE]

Using the same approach as for concurrence, we have the following Theorems.

Theorem 7**.**

Suppose $k$ is a real number satisfying $0<k\leq 1$. For any $N$-qubit mixed state $\rho_{AB_{1}\cdots B_{N-1}}$, if $kE_{AB_{i}}^{a}\geq E_{A|B_{i+1}\cdots B_{N-1}}^{a}$ for $i=1,2,\ldots,m$, and $E_{AB_{j}}^{a}\leq kE_{A|B_{j+1}\cdots B_{N-1}}^{a}$ for $j=m+1,\ldots,N-2$, $\forall 1\leq m\leq N-3$, $N\geq 4$, we have

[TABLE]

for $0\leq\beta\leq 1$.

Theorem 8**.**

Suppose $k$ is a real number satisfying $0<k\leq 1$. For any $N$-qubit mixed state $\rho_{AB_{1}\cdots B_{N-1}}$, if $kE_{AB_{i}}^{a}\geq E_{A|B_{i+1}\cdots B_{N-1}}^{a}$ for all $i=1,2,\ldots,N-2$, we have

[TABLE]

for $0\leq\beta\leq 1$.

IV Tighter constraints related to negativity

The negativity, a well-known quantifier of bipartite entanglement, is defined as \mathcal{N}(\rho_{AB})=\big{(}\|\rho_{AB}^{T_{A}}\|-1\big{)}/2 Vidal , where $\rho_{AB}^{T_{A}}$ is the partial transposed matrix of $\rho_{AB}$ with respect to the subsystem $A$, and $\|X\|$ denotes the trace norm of $X$, i.e., $\|X\|={\rm tr}\sqrt{XX^{{\dagger}}}$. For convenient, we use the definition of negativity as $\|\rho_{AB}^{T_{A}}\|-1$. Particularly, for any bipartite pure state $|\psi\rangle_{AB}$, $\mathcal{N}(|\psi\rangle_{AB})=2\sum\limits_{i<j}\sqrt{\lambda_{i}\lambda_{j}}=({\rm tr}\sqrt{\rho_{A}})^{2}-1$, where $\lambda_{i}s$ are the eigenvalues of the reduced density matrix $\rho_{A}={\rm tr}_{B}|\psi\rangle_{AB}\langle\psi|$. The convex-roof extended negativity (CREN) of a mixed state $\rho_{AB}$ is defined by

[TABLE]

where the minimum is taken over all possible pure state decomposition of $\rho_{AB}$. Thus $\mathcal{N}_{c}(\rho_{AB})=\mathcal{C}(\rho_{AB})$ for any two-qubit mixed state $\rho_{AB}$. The dual to the CREN of a mixed state $\rho_{AB}$ is defined as

[TABLE]

with the maximum taking over all possible pure state decomposition of $\rho_{AB}$. Furthermore, $\mathcal{N}^{a}_{c}(\rho_{AB})=\mathcal{C}^{a}(\rho_{AB})$ for any two-qubit mixed state $\rho_{AB}$JSK8 .

Similar to the concurrence and EoF, we have the following Theorems.

Theorem 9**.**

Suppose $k$ is a real number satisfying $0<k\leq 1$. For any $N$-qubit mixed state $\rho_{AB_{1}\cdots B_{N-1}}$, if $k\mathcal{N}_{cAB_{i}}^{2}\geq\mathcal{N}_{cA|B_{i+1}\cdots B_{N-1}}^{2}$ for $i=1,2,\ldots,m$, and $\mathcal{N}_{cAB_{j}}^{2}\leq k\mathcal{N}_{cA|B_{j+1}\cdots B_{N-1}}^{2}$ for $j=m+1,\ldots,N-2$, $\forall 1\leq m\leq N-3$, $N\geq 4$, then we have

[TABLE]

for all $\beta\geq 2$.

Theorem 10**.**

Suppose $k$ is a real number satisfying $0<k\leq 1$. For any $N$-qubit mixed state $\rho_{AB_{1}\cdots B_{N-1}}$, if all $k\mathcal{N}_{cAB_{i}}^{2}\geq\mathcal{N}_{cA|B_{i+1}\cdots B_{N-1}}^{2}$ for all $i=1,2,\ldots,N-2$, then

[TABLE]

for $\beta\geq 2$.

Example 3    Consider the state in Example 1 with $\lambda_{0}=\lambda_{3}=\frac{1}{2},\ \lambda_{2}=\frac{\sqrt{2}}{2}$ and $\lambda_{1}=\lambda_{4}=0$. We have $\mathcal{N}_{cA|BC}=\frac{\sqrt{3}}{2}$, $\mathcal{N}_{cAB}=\frac{\sqrt{2}}{2}$ and $\mathcal{C}_{cAC}=\frac{1}{2}$. Then \mathcal{N}_{cAB}^{\beta}+\big{(}2^{\frac{\beta}{2}}-1\big{)}\mathcal{N}_{cAC}^{\beta}=\big{(}\frac{\sqrt{2}}{2}\big{)}^{\beta}+\big{(}2^{\frac{\beta}{2}}-1\big{)}\big{(}\frac{1}{2}\big{)}^{\beta} and \mathcal{N}_{cAB}^{\beta}+\frac{(1+k)^{\frac{\beta}{2}}-1}{k^{\frac{\beta}{2}}}\mathcal{N}_{cAC}^{\beta}=\big{(}\frac{\sqrt{2}}{2}\big{)}^{\beta}+\frac{(1+k)^{\frac{\beta}{2}}-1}{k^{\frac{\beta}{2}}}\big{(}\frac{1}{2}\big{)}^{\beta}. One can see that our result is better than the one in SM.Fei3 for $\beta\geq 2$, thus also better than the ones in SM.Fei1 ; SM.Fei2 , see Fig. 3.

For the negativity of assistance $\mathcal{N}_{c}^{a}$, we have the following results.

Theorem 11**.**

Suppose $k$ is a real number satisfying $0<k\leq 1$. For an $N$-qubit pure state $|\psi\rangle_{A|B_{1}\cdots B_{N-1}}$, if $k(\mathcal{N}^{a}_{cA|B_{i}})^{2}\geq(\mathcal{N}^{a}_{cA|B_{i+1}\cdots B_{N-1}})^{2}$ for $i=1,2,\ldots,m$, and $(\mathcal{N}^{a}_{cAB_{j}})^{2}\leq k(\mathcal{N}^{a}_{cA|B_{j+1}\cdots B_{N-1}})^{2}$ for $j=m+1,\ldots,N-2$, $\forall 1\leq m\leq N-3$, $N\geq 4$, then we have

[TABLE]

for all $0\leq\beta\leq 2$.

Theorem 12**.**

Suppose $k$ is a real number satisfying $0<k\leq 1$. For any $N$-qubit mixed state $|\psi\rangle_{AB_{1}\cdots B_{N-1}}$, if $k(\mathcal{N}^{a}_{cAB_{i}})^{2}\geq(\mathcal{N}^{a}_{cA|B_{i+1}\cdots B_{N-1}})^{2}$ for all $i=1,2,\ldots,N-2$, then

[TABLE]

for $0\leq\beta\leq 2$.

V Tighter monogamy relations for Tsallis-$q$ entanglement and Rényi-$\alpha$ entanglement

In this section, we study the Tsallis-$q$ entanglement and Rényi-$\alpha$ entanglement, and establish the corresponding monogamy and polygamy relations for the two entanglement measures, respectively.

V.1 Tighter monogamy and polygamy relations for Tsallis-$q$ entanglement

The Tsallis-$q$ entanglement of a bipartite pure state $|\psi\rangle_{AB}$ is defined as JSK3

[TABLE]

where $q>0$ and $q\neq 1$. For the case $q$ tends to 1, $T_{q}(\rho)$ is just the von Neumann entropy, $\lim\limits_{q\rightarrow 1}T_{q}(\rho)=-{\rm tr}\rho\log_{2}\rho=S(\rho)$. The Tsallis-$q$ entanglement of a bipartite mixed state $\rho_{AB}$ is given by $T_{q}(\rho_{AB})=\min\limits_{\{p_{i},|\psi_{i}\rangle\}}\sum\limits_{i}p_{i}T_{q}(|\psi_{i}\rangle)$ with the minimum taken over all possible pure state decompositions of $\rho_{AB}$. For $\frac{5-\sqrt{13}}{2}\leq q\leq\frac{5+\sqrt{13}}{2}$, Yuan et al. proposed an analytic relationship between the Tsallis-$q$ entanglement and concurrence,

[TABLE]

where

[TABLE]

with $0\leq x\leq 1$ Yuan . It has also been proved that $T_{q}(|\psi\rangle)=g_{q}(\mathcal{C}^{2}(|\psi\rangle))$ if $|\psi\rangle$ is a $2\otimes m$ pure state, and $T_{q}(\rho)=g_{q}(\mathcal{C}^{2}(\rho))$ if $\rho$ is a two-qubit mixed state. Hence, (42) holds for any $q$ such that $g_{q}(x)$ in (43) is monotonically increasing and convex. Particularly, one has that

[TABLE]

for $2\leq q\leq 3$. In Ref. JSK3 , Kim provided a monogamy relation for the Tsallis-$q$ entanglement,

[TABLE]

where $i=1,2,\ldots,N-1$ and $2\leq q\leq 3$. Later, this relation was improved as follows: if $\mathcal{C}_{AB_{i}}\geq\mathcal{C}_{A|B_{i+1}\cdots B_{N-1}}$ for $i=1,2,\ldots,m$, and $\mathcal{C}_{AB_{j}}\leq\mathcal{C}_{A|B_{j+1}\cdots B_{N-1}}$ for $j=m+1,\ldots,N-2$, $\forall 1\leq m\leq N-3$, $N\geq 4$, then

[TABLE]

where $\beta\geq 1$ and $T^{\beta}_{qA|B_{1}B_{2}\cdots B_{N-1}}$ quantifies the Tsallis-$q$ entanglement under partition $A|B_{1}B_{2}\cdots B_{N-1}$, and $T^{\beta}_{qA|B_{i}}$ quantifies that of the two-qubit subsystem $AB_{i}$ with $2\leq q\leq 3$. Moreover, for $\frac{5-\sqrt{13}}{2}\leq q\leq\frac{5+\sqrt{13}}{2}$, one has

[TABLE]

We now provide monogamy relations which are tighter than (45) and (46).

Theorem 13**.**

Suppose $k$ is a real number satisfying $0<k\leq 1$. For an arbitrary $N$-qubit mixed state $\rho_{AB_{1}B_{2}\cdots B_{N-1}}$, if $kT_{qAB_{i}}\geq T_{qA|B_{i+1}\cdots B_{N-1}}$ for $i=1,2,\ldots,m$, and $T_{qAB_{j}}\leq kT_{qA|B_{j+1}\cdots B_{N-1}}$ for $j=m+1,\ldots,N-2$, $\forall 1\leq m\leq N-3$, $N\geq 4$, then we have

[TABLE]

for $\beta\geq 1$ and $2\leq q\leq 3$.

Theorem 14**.**

Suppose $k$ is a real number satisfying $0<k\leq 1$. For any $N$-qubit mixed state $\rho_{AB_{1}\cdots B_{N-1}}$, if all $kT_{qAB_{i}}\geq T_{qA|B_{i+1}\cdots B_{N-1}}$ for $i=1,2,\ldots,N-2$, then we have

[TABLE]

for $\beta\geq 1$ and $2\leq q\leq 3$.

Example 4    Consider the quantum state given in Example 1 with $\lambda_{0}=\lambda_{3}=\frac{1}{2},\ \lambda_{2}=\frac{\sqrt{2}}{2}$ and $\lambda_{1}=\lambda_{4}=0$. For $q=2$, one has $T_{2A|BC}=\frac{3}{8}$, $T_{2AB}=\frac{1}{4}$ and $T_{2AC}=\frac{1}{8}$. Then T_{2AB}^{\beta}+\big{(}2^{\beta}-1\big{)}T_{2AC}^{\beta}=(\frac{1}{4})^{\beta}+\big{(}2^{\beta}-1\big{)}(\frac{1}{8})^{\beta} and T_{2AB}^{\beta}+\frac{(1+k)^{\beta}-1}{k^{\beta}}T_{2AC}^{\beta}=\big{(}\frac{1}{4}\big{)}^{\beta}+\frac{(1+k)^{\beta}-1}{k^{\beta}}\big{(}\frac{1}{8}\big{)}^{\beta}. It can be seen that our result is better than the one in SM.Fei3 for $\beta\geq 1$, and also better than the ones given in SM.Fei1 ; SM.Fei2 , see Fig. 4.

As a dual quantity to Tsallis-$q$ entanglement, the Tsallis-$q$ entanglement of assistance (TEoA) is defined by JSK3 , $T_{q}^{a}(\rho_{AB})=\max\limits_{\{p_{i},|\psi_{i}\rangle\}}\sum\limits_{i}p_{i}T_{q}(|\psi_{i}\rangle)$, where the maximum is taken over all possible pure state decompositions of $\rho_{AB}$. If $1\leq q\leq 2$ or $3\leq q\leq 4$, the function $g_{q}$ defined in (43) satisfies

[TABLE]

which leads to the Tsallis polygamy inequality

[TABLE]

for any multi-qubit state $\rho_{A|B_{1}B_{2}\cdots B_{N-1}}$ JSK9 . Here we provide tighter polygamy relations related to Tsallis-$q$ entanglement. We have the following results.

Theorem 15**.**

Suppose $k$ is a real number satisfying $0<k\leq 1$. For any $N$-qubit mixed state $\rho_{AB_{1}\cdots B_{N-1}}$, if $kT^{a}_{qAB_{i}}\geq T^{a}_{qA|B_{i+1}\cdots B_{N-1}}$ for $i=1,2,\ldots,m$, and $T^{a}_{qAB_{j}}\leq kT^{a}_{qA|B_{j+1}\cdots B_{N-1}}$ for $j=m+1,\ldots,N-2$, $\forall 1\leq m\leq N-3$, $N\geq 4$, then

[TABLE]

for $0\leq\beta\leq 1$ with $1\leq q\leq 2$ or $3\leq q\leq 4$.

Theorem 16**.**

Suppose $k$ is a real number satisfying $0<k\leq 1$. For any $N$-qubit mixed state $\rho_{AB_{1}\cdots B_{N-1}}$, if $kT^{a}_{qAB_{i}}\geq T^{a}_{qA|B_{i+1}\cdots B_{N-1}}$ for all $i=1,2,\ldots,N-2$, we have

[TABLE]

for $0\leq\beta\leq 1$ with $1\leq q\leq 2$ or $3\leq q\leq 4$.

V.2 Tighter monogamy and polygamy relations for Rényi-$\alpha$ entanglement

For a bipartite pure state $|\psi\rangle_{AB}$, the Rényi-$\alpha$ entanglement is defined as vidal $E(|\psi\rangle_{AB})=S_{\alpha}(\rho_{A})$, where $S_{\alpha}(\rho)=\frac{1}{1-\alpha}\log_{2}{\rm tr}\rho^{\alpha}$ for any $\alpha>0$ and $\alpha\neq 1$, and $\lim\limits_{\alpha\rightarrow 1}S_{\alpha}(\rho)=S(\rho)=-{\rm tr}\rho\log_{2}\rho$. For a bipartite mixed state $\rho_{AB}$, the Rényi-$\alpha$ entanglement is given by

[TABLE]

where the minimum is taken over all possible pure-state decompositions of $\rho_{AB}$. For each $\alpha>0$, one has $E_{\alpha}(\rho_{AB})=f_{\alpha}(\mathcal{C}(\rho_{AB}))$, where f_{\alpha}(x)=\frac{1}{1-\alpha}\log\big{[}\big{(}\frac{1-\sqrt{1-x^{2}}}{2}\big{)}^{2}+(\frac{1+\sqrt{1-x^{2}}}{2}\big{)}^{2}\big{]} is a monotonically increasing and convex function JSK2 . For $\alpha\geq 2$ and any $n$-qubit state $\rho_{A|B_{1}B_{2}\cdots B_{N-1}}$, one has JSK3

[TABLE]

We propose the following two monogamy relations for the Rényi-$\alpha$ entanglement, which are tighter than the previous results.

Theorem 17**.**

Suppose $k$ is a real number satisfying $0<k\leq 1$. For an arbitrary $N$-qubit mixed state $\rho_{AB_{1}B_{2}\cdots B_{N-1}}$, if $kE_{\alpha AB_{i}}\geq E_{\alpha A|B_{i+1}\cdots B_{N-1}}$ for $i=1,2,\ldots,m$, and $E_{\alpha AB_{j}}\leq kT_{\alpha A|B_{j+1}\cdots B_{N-1}}$ for $j=m+1,\ldots,N-2$, $\forall 1\leq m\leq N-3$, $N\geq 4$, then

[TABLE]

for $\beta\geq 1$ and $\alpha\geq 2$.

Theorem 18**.**

Suppose $k$ is a real number satisfying $0<k\leq 1$. For an arbitrary $N$-qubit mixed state $\rho_{AB_{1}B_{2}\cdots B_{N-1}}$, if $kE_{\alpha AB_{i}}\geq E_{\alpha A|B_{i+1}\cdots B_{N-1}}$ for all $i=1,2,\ldots,N-2$, then

[TABLE]

for $\beta\geq 1$ and $\alpha\geq 2$.

Example 5    Consider again the state given in Example 1 with $\lambda_{0}=\lambda_{3}=\frac{1}{2},\ \lambda_{2}=\frac{\sqrt{2}}{2}$ and $\lambda_{1}=\lambda_{4}=0$. For $\alpha=2$, we find $E_{2A|BC}=\log_{2}\frac{8}{5}\approx 678072$, $E_{2AB}=\log_{2}\frac{8}{7}\approx 0.415037$ and $E_{2AC}=\log_{2}\frac{4}{3}\approx 0.192645$. Then $E_{2AB}^{\alpha}+E_{2AC}^{\alpha}=0.415037^{\alpha}+0.192645^{\alpha}$ and $E_{2AB}^{\alpha}+\frac{(1+k)^{\alpha}-1}{k^{\alpha}}E_{2AC}^{\alpha}=0.415037^{\alpha}+\frac{(1+k)^{\alpha}-1}{k^{\alpha}}0.192645^{\alpha}$. One can see that our result is better than the result in JSK3 , and the smaller $k$ is, the tighter relation is, see Fig. 5.

The Rényi-$\alpha$ entanglement of assistance (REoA), a dual quantity to Rényi-$\alpha$ entanglement, is defined as $E_{\alpha}^{a}(\rho_{AB})=\max\limits_{\{p_{i},|\psi_{i}\rangle\}}\sum\limits_{i}p_{i}E_{\alpha}(|\psi_{i}\rangle)$, where the maximum is taken over all possible pure state decompositions of $\rho_{AB}$. For $\alpha\in[\frac{\sqrt{7}-1}{2},\frac{\sqrt{13}-1}{2}]$ and any $n$-qubit state $\rho_{AB_{1}B_{2}\cdots B_{N-1}}$, a polygamy relation of multi-partite quantum entanglement in terms of REoA has been given by Song :

[TABLE]

We improve this inequality to be a tighter ones under some netural conditions.

Theorem 19**.**

Suppose $k$ is a real number satisfying $0<k\leq 1$. For an arbitrary $N$-qubit mixed state $\rho_{AB_{1}B_{2}\cdots B_{N-1}}$, if $kE^{a}_{\alpha AB_{i}}\geq E^{a}_{\alpha A|B_{i+1}\cdots B_{N-1}}$ for $i=1,2,\ldots,m$, and $E^{a}_{\alpha AB_{j}}\leq kE^{a}_{\alpha A|B_{j+1}\cdots B_{N-1}}$ for $j=m+1,\ldots,N-2$, $\forall 1\leq m\leq N-3$, $N\geq 4$, then

[TABLE]

for $0\leq\beta\leq 1$ with $\frac{\sqrt{7}-1}{2}\leq\alpha\leq\frac{\sqrt{13}-1}{2}$.

Theorem 20**.**

Suppose $k$ is a real number satisfying $0<k\leq 1$. For an arbitrary $N$-qubit mixed state $\rho_{AB_{1}B_{2}\cdots B_{N-1}}$, if $kE^{a}_{\alpha AB_{i}}\geq E^{a}_{\alpha A|B_{i+1}\cdots B_{N-1}}$ for all $i=1,2,\ldots,N-2$, then

[TABLE]

for $0\leq\beta\leq 1$, with $\frac{\sqrt{7}-1}{2}\leq\alpha\leq\frac{\sqrt{13}-1}{2}$.

VI Conclusion

Both entanglement monogamy and polygamy are fundamental properties of multipartite entangled states. We have presented monogamy relations related to the $\beta$th power of concurrence, entanglement of formation, negativity, Tsallis-$q$ and Rényi-$\alpha$ entanglement. We also provide polygamy relations related to these entanglement measures. All the relations we presented in this paper are tighter than the previous results. These tighter monogamy and polygamy inequalities can also provide finer characterizations of the entanglement distributions among the multiqubit systems. Our results provide a rich reference for future work on the study of multiparty quantum entanglement. And our approaches are also useful for further study on the monogamy and polygamy properties related to measures of other quantum correlations and quantum coherence framework ; coherence .

VII Acknowledgements

This work is supported by the National Natural Science Foundation of China under Grant Nos. 11805143 and 11675113, and Key Project of Beijing Municipal Commission of Education under No. KZ201810028042.